a
Here, \(C_{1}=\frac{2 \varepsilon_{0} k_{1} A}{3 d}, C_{2}=\frac{2 \varepsilon_{0} k_{2} A}{3 d}\)

\(C_{3}=\frac{2 \varepsilon_{0} k_{3} A}{3 d}, C_{4}=\frac{2 \varepsilon_{0} k_{4} A}{d}\)

Given system of \(C_{1}, C_{2}, C_{3}\) and \(C_{4}\) can be simplified as \(\therefore \frac{1}{C_{A B}}=\frac{1}{C_{1}+C_{2}+C_{3}}+\frac{1}{C_{4}}\)

Suppose, \(C_{A B}=\frac{k \varepsilon_{0} A}{d}\)

\(\quad \frac{1}{k\left(\frac{\varepsilon_{0} A}{d}\right)}=\frac{1}{\frac{2}{3} \frac{\varepsilon_{0} A}{d}\left(k_{1}+k_{2}+k_{3}\right)}+\frac{1}{\frac{2 \varepsilon_{0} A}{d} k_{4}}\)

\(\Rightarrow \frac{1}{k}=\frac{3}{2\left(k_{1}+k_{2}+k_{3}\right)}+\frac{1}{2 k_{4}} \therefore \frac{2}{k}=\frac{3}{k_{1}+k_{2}+k_{3}}+\frac{1}{k_{4}}\)